I've read there are two admissibility criteria for wavelets, both of which are designed to preserve total power of the signal (source: http://en.wikipedia.org/wiki/Wavelet#Mother_wavelet, as well as various scientific papers)

1. Condition for zero-mean: $$\int_{-\infty}^{\infty}\psi(t)dt=0$$
2. Condition for square norm one: $$\int_{-\infty}^{\infty}|\psi(t)|^2dt=1,$$

where $\psi(t)$ is the wavelet kernel. This leads to having a normalization factor of $\frac{1}{\sqrt{2\pi}}$ for the Gabor/Morlet wavelet.

Now, my question is: how do these admissibility criteria apply to 2D wavelets? (Let's say $\psi(x,y)$.)

My guess would be:

1. $$\int_{x=-\infty}^{\infty}\int_{y=-\infty}^{\infty}\psi(x,y)dx dy=0$$
2. $$\int_{x=-\infty}^{\infty}\int_{y=-\infty}^{\infty}|\psi(x,y)|^2dxdy=1,$$

but I get inconsistent results in my computations.

Is my guess right? If not, what are they? Can you provide a reliable source?

• What exactly is being inconsistent in your computations? From what I understand your admissibility criterions are correct, but I will have to double check when I get home. Still, what is inconsistent in your calculations? – Spacey Jan 29 '13 at 23:31
• When comparing wavelets coefficients to Fourier coefficients, I'm off by about 5-6 orders of magnitude (beside that, everything is fine: the slope and overall shape is good). It's just a matter of getting my code straight (which is a completely different problem!) If I can just confirm my normalisation constant is good, it'll save me a lot of time! – PhilMacKay Jan 30 '13 at 14:43

It turns out the admissibility criterion does not dictate that the energy of the wavelet must be unity. To be classified as a wavelet, a wavelet must follow the following criteria:

• The wavelet must have finite energy, so $E = \int_{-\infty}^{+\infty} |\psi(t)|^{2} dt < \infty$
• The second condition is that the wavelet must have zero mean. (The intuition behind this is because the wavelet is acting like a matched filter, and we care only about how nicely the shape of the received signal matches the wavelet, and not the energy of the received signal). So if $\hat\psi(f)$ is the fourier transform of your wavelet, then: $$C_g = \int_{0}^{\infty} \frac{|\hat\psi(f)|^2}{f} df < \infty$$ The $C_g$ here is called the admissibility constant, and the above property is called the admissibility criterion. This implies that $\hat\psi(0)=0$, because if it didn't, the above integral would blow up as you can see.
• Finally, for complex wavelets, the last criterion states that the Fourier transform must be both real, and vanish for negative frequencies.

Now notice, no where does it state that the square norm of the wavelet must be equal to unity. It only states that the $E < \infty$.

In your case, it sounds as though the energy of your signal in the Wavelet domain, is not matching the energy of your signal in the Fourier domain. My best guess without further information is that you not normalize your wavelet to unit energy, since it is not demanded by the admissibility criterion. (First check though, if the energy of your signal in the spatial domain actually matches the energy of your signal in the wavelet domain). I would start there.

• Accepted, for answering my question in details! Thanks! It turns out my specific application requires carefull normalization: dx.doi.org/10.1016/j.cageo.2005.01.014 See section 5 for details, note that he also uses a different (but very similar) standard for admissibility (Section 3). – PhilMacKay Jan 30 '13 at 19:20
• @PhilMacKay There also appears to be a typo in the paper: Equation (5) states that "they must have finite energy", and then goes on to say that: $$\int\int_{-\infty}^{\infty}|\psi(\bf{x})|^2d^2x = \int\int_{-\infty}^{\infty}|\hat\psi(k)|^2d^2k =0$$ It seems as though it should read "$< \infty$" instead of "$=0$" – Spacey Jan 30 '13 at 21:58
• @Mohammad One other thing I wanted to share on the topic of Wavelets is a website written by an practicing engineer, aimed at the "engineering mine" that covers wavelet theory in a "user friendly way. polyvalens.com/blog – user2718 Feb 1 '13 at 18:22
• @BruceZenone Thanks again Bruce, Ill have to add that to my ever expanding list! :-) – Spacey Feb 3 '13 at 22:51
• @Mohammad I did notice this odd definition... In practice, I usually normalise such that it equals unity, because then the result is of the same order of magnitude as the initial data. – PhilMacKay Feb 5 '13 at 16:42